AQA Further AS Paper 2 Discrete 2023 June — Question 7 7 marks

Exam BoardAQA
ModuleFurther AS Paper 2 Discrete (Further AS Paper 2 Discrete)
Year2023
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMinimum Spanning Trees
TypeMaximize spanning tree weight
DifficultyStandard +0.3 This is a straightforward application of maximum spanning tree algorithm (Kruskal's/Prim's in reverse) on a small network with clear arc weights. The twist of maximizing rather than minimizing is trivial (just select longest edges instead of shortest). Part (b) is basic proportion calculation, and part (c) requires a simple real-world modeling comment. Slightly easier than average due to being a standard algorithm application with no conceptual challenges.
Spec7.04b Minimum spanning tree: Prim's and Kruskal's algorithms

7 A construction company has built eight wind turbines on a moorland site. The network below shows nodes which represent the site entrance, \(E\), and the wind turbine positions, \(S , T , \ldots , Z\) \includegraphics[max width=\textwidth, alt={}, center]{372edcfa-c3cd-4c83-89e9-2bb5fd9825f1-12_924_1294_479_356} Each arc represents an access track with its length given in metres.
These 17 tracks were created in order to build the wind turbines. Eight of the tracks are to be retained so that each turbine can be accessed for maintenance, directly or indirectly, from the site entrance. The other nine tracks will be removed. 7
    1. To save money the construction company wants to maximise the total length of the eight tracks to be retained. Determine which tracks the construction company should retain.
      7
      1. (ii) Find the total length of the eight tracks that are to be retained. 7
    2. The total length of the 17 tracks is 14.6 km
      The cost of removing all 17 tracks would be \(\pounds 87,600\) Using your answer to part (a)(ii), calculate an estimate for the cost of removing the nine tracks that will not be retained.
      [0pt] [2 marks]
      7
    3. Comment on why the modelling used in part (b) may not give an accurate estimate for the cost of removing the nine tracks.

Question 7(a)(i):
AnswerMarks Guidance
AnswerMarks Guidance
Lists 8 arcs: \(ET\): 1000, \(EV\): 990, \(ES\): 960, \(EZ\): 920, \(TU\): 910, \(UX\): 900, \(XY\): 890, \(YW\): 850M1 Sets up a model of finding a maximum spanning tree by listing 8 arcs with at least 4 correct
Correct 8 arcs of the maximum spanning tree (see diagram)A1 Uses their model to find the 8 correct arcs of the maximum spanning tree
Question 7(a)(ii):
AnswerMarks Guidance
AnswerMarks Guidance
\(1000 + 990 + 960 + 920 + 910 + 900 + 890 + 850 = 7420\) metresM1 Finds the total of the weights of the 8 arcs from their model
\(7420\) metresA1F Finds the correct length with units from their model
Question 7(b):
AnswerMarks Guidance
Average cost of track removal is \(87600 \div 14.6 = \pounds6000\) per kmM1 (AO 3.1a) Calculates an estimate of cost by using an average cost
Need to remove \((14.6 - 7.42)\) km of track
AnswerMarks Guidance
\((14.6 - 7.42) \times 6000 = \pounds43,080\)A1F (AO 3.2a) Calculates the correct estimate for the total cost, with units, for their total length
Question 7(c):
AnswerMarks Guidance
The cost of track removal may not be proportional to track length.B1 (AO 3.5b) Comments on the limitation of modelling the removal cost per unit length with an average cost
# Question 7(a)(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Lists 8 arcs: $ET$: 1000, $EV$: 990, $ES$: 960, $EZ$: 920, $TU$: 910, $UX$: 900, $XY$: 890, $YW$: 850 | M1 | Sets up a model of finding a maximum spanning tree by listing 8 arcs with at least 4 correct |
| Correct 8 arcs of the maximum spanning tree (see diagram) | A1 | Uses their model to find the 8 correct arcs of the maximum spanning tree |

---

# Question 7(a)(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $1000 + 990 + 960 + 920 + 910 + 900 + 890 + 850 = 7420$ metres | M1 | Finds the total of the weights of the 8 arcs from their model |
| $7420$ metres | A1F | Finds the correct length with units from their model |

## Question 7(b):

Average cost of track removal is $87600 \div 14.6 = \pounds6000$ per km | M1 (AO 3.1a) | Calculates an estimate of cost by using an average cost

Need to remove $(14.6 - 7.42)$ km of track

$(14.6 - 7.42) \times 6000 = \pounds43,080$ | A1F (AO 3.2a) | Calculates the correct estimate for the total cost, with units, for their total length

---

## Question 7(c):

The cost of track removal may not be proportional to track length. | B1 (AO 3.5b) | Comments on the limitation of modelling the removal cost per unit length with an average cost

---
7 A construction company has built eight wind turbines on a moorland site.

The network below shows nodes which represent the site entrance, $E$, and the wind turbine positions, $S , T , \ldots , Z$\\
\includegraphics[max width=\textwidth, alt={}, center]{372edcfa-c3cd-4c83-89e9-2bb5fd9825f1-12_924_1294_479_356}

Each arc represents an access track with its length given in metres.\\
These 17 tracks were created in order to build the wind turbines.

Eight of the tracks are to be retained so that each turbine can be accessed for maintenance, directly or indirectly, from the site entrance.

The other nine tracks will be removed.

7
\begin{enumerate}[label=(\alph*)]
\item (i) To save money the construction company wants to maximise the total length of the eight tracks to be retained.

Determine which tracks the construction company should retain.\\

7 (a) (ii) Find the total length of the eight tracks that are to be retained.

7
\item The total length of the 17 tracks is 14.6 km\\
The cost of removing all 17 tracks would be $\pounds 87,600$\\
Using your answer to part (a)(ii), calculate an estimate for the cost of removing the nine tracks that will not be retained.\\[0pt]
[2 marks]\\

7
\item Comment on why the modelling used in part (b) may not give an accurate estimate for the cost of removing the nine tracks.
\end{enumerate}

\hfill \mbox{\textit{AQA Further AS Paper 2 Discrete 2023 Q7 [7]}}